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Electric field due to an electric dipole on its - Consider, an electric dipole of length 2a having charges +q and  -q. A point P lies on the equatorial line of the dipole at a distance r from the centre of the dipole. The electric fields due to '-q' and '+q' are E1 and E2 respectively. Now, | E1 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(1) | E2 |  = kq/[(r² + a²)½]² = kq/(r² + a²)    .....(2) From equation (1) and (2) | E1 |  =  | E2 |  =   x (say) Clearly, components E1sinθ and E2sinθ cancel each other. Therefore, Net electric field at point P is - E = xCosθ + xCosθ = 2xCosθ From equation (1) E = 2kq.Cosθ/(r² + a²)         = kq2a/[(r² + a²).(r² + a²)½]   = kq2a/[(r² + a²)³]½ Also, dipole moment - Questions for Practice Q1. Find the ratio of electric fields due to a short dipole on the axial line to equatorial line at same distances from the centre. Ans.    Ea/Ee = (2kp/r³)/(kp/r³)                   = 2

Consider, an electric dipole of length 2a having charges +q and -q . A point P is at a distance r from the centre of the dipole such that OP is making an angle θ with dipole moment p . CONSTRUCTION :- Extend OP to BP such that BC is perpendicular to BP, Draw AD perpendicular to BP. PROOF :- Now, β = r + aCosθ      .........(1)          α = r + aCosθ      .........(2) Let, 'V1' and 'V2' be the potential due to +q and -q respectively. Net electric potential at 'P' is -               V = V1 + V2             V = kq/α + k(-q)/β             From equation (1) and (2)            Also,       p = q.2a Therefore,               [  V = kpCosθ/(r² - a²Cos²θ)   ] Note :- For axial line, θ = 0           For equatorial line, θ = π/2 Pyaare bacchon, tumhari help k liye video daali hai by  www.advguruji.blogspot.com

Consider, an electric dipole having dipole moment p in an uniform electric field E such that the dipole moment is making an angle θ with the field. Now, small amount of work done dW in rotating the dipole by small angle dθ is - Total work done when dipole is rotated from angle θ1 to angle θ2 with the field is -                             =  -pE[Cosθ2 - Cosθ1]                        W =  pE[Cosθ1 - Cosθ2] This work done will be stored as potential energy of the dipole. Therefore,  U =  pE[Cosθ1 - Cosθ2] NOTE :- As we know, more the potential energy of a system, less it will be stable and vice versa. 1. For most stable condition    [U = minimum]        θ1 = 180° ,    θ2 = 0° Therefore, U = -2pE 2. For most unstable condition    [U = maximum]       θ1 = 0°  ,     θ2 = 180° Therefore, U = pE[1 - (-1)]                    U = 2pE Still not clear click here to watch      by  www.advguruji.blogspot.com

Consider, an electric dipole of length '2a' having charges '+q' and '-q' placed in an uniform electric field 'E'. The dipole moment 'p' is making an angle 'θ' with the electric field. There are two equal amount of electrostatic forces acting in opposite direction on the dipole which resulting zero net force but a torque 'τ' on the dipole. As, torque = force × perpendicular distance          τ = qE × 2aSinθ For "potential energy of an electric dipole rotated in an uniform electric field" CLICK HERE by      www.advguruji.blogspot.com

Consider, a short electric dipole having dipole moment p . A point P is at a distance r from the centre O of the dipole such that the angle between p and OP is θ . Now, P act as a point on the axial line for pCosθ component and act as a point on the equatorial line for pSinθ component. The electric fields at point P due to pCosθ and pSinθ are E1 and E2 respectively. Also, they are perpendicular to each other. Now, net electric field at point P is - by www.advguruji.blogspot.com

Consider, an electric dipole of length 2a having charges +q and -q . A point P is lying on the axial line of the dipole at distance r from the centre of the dipole (as shown in figure). The electric fields due to charges -q and +q are E1 and E2 respectively. for Electric field due to an electric dipole on its equatorial line CLICK HERE by     www.advguruji.blogspot.com

Electric dipole If two equal and opposite charges placed at some distance apart. This two charge system is called an electric dipole. Here, '2a' is the dipole length. Note :- Total charge in an electric dipole = -q + q = 0 Electric dipole moment 'p' It is the product of the magnitude of one charge of the dipole and the dipole length. It is a vector quantity directed from negative charge to positive charge. Its SI unit is Cm(coulomb metre). for Electric field due to an electric dipole on its axial line  CLICK HERE   by         www.advguruji.blogspot.com